2008-II-21
Ans: D
Since $ABCD$ is a parallelogram, then $AD = BC$. Hence
… Read Since $ABCD$ is a parallelogram, then $AD = BC$. Hence
$\begin{array}{rcl}
\tan 30
Note that
$\begin{array}{rcl}
AC^2 + BC^2 & =
Let $\angle ADC = y$. Add a point $E$ on $AB$ such that $E
Add a horizontal line and the line divides $b$ into $b_
According to the above figure, we have
$\begin{array
$\begin{array}{cl}
=
$\begin{array}{cl}
= &